Animated Solution for Physics - Current Electricity: A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances etc, is shown in the figure. Calculate the energy stored in the capacitor $C$ ($4\mu \text{F}$).

The problem presents us with a fascinating scenario: a snapshot of a larger, complex circuit in its steady state. Our mission is to find the energy stored in the $4\mu \text{F}$ capacitor. To do this, we must act like electrical detectives, tracing the currents and finding a valid path to determine the potential difference across the capacitor.

The first clue lies in the phrase "steady state". When a DC circuit reaches its steady state, the capacitors are fully charged. They act as open switches, meaning absolutely no current flows through the branch containing the capacitor.

Imagine the capacitor as a dam that has completely filled up; water (current) can no longer flow through that path. Therefore, the current flowing from node $a$ to node $b$ through the capacitor is $0\text{ A}$.

Now, let's apply Kirchhoff's Current Law (KCL), which states that the total current entering a junction must equal the total current leaving it.

At Junction $a$: - Current entering from the left branch = $2\text{ A}$ - Current entering from the top branch = $1\text{ A}$ - Total current entering = $3\text{ A}$

Since no current can go down through the capacitor, all $3\text{ A}$ must exit to the right. Thus, the current flowing through the $5\Omega$ resistor is $3\text{ A}$ towards the right.

At Junction $b$: - Current entering from the left branch = $2\text{ A}$ - Current entering from the bottom branch = $1\text{ A}$ - Total current entering = $3\text{ A}$

Again, no current comes from the capacitor branch. Therefore, all $3\text{ A}$ must exit to the right. The current flowing through the $2\Omega$ resistor is $3\text{ A}$ towards the right.

Following the $3\text{ A}$ current past the $5\Omega$ resistor to junction $c$, the circuit diagram implies this entire current flows downwards through the $1\Omega$ resistor to reach junction $d$. This means the current through the $1\Omega$ resistor is $3\text{ A}$ downwards.

Choosing the Right Path

To find the energy stored in the capacitor, we need the potential difference across it, $V_{ab}=V_a-V_b$.

Here is where many students fall into a trap. The problem explicitly states this is "A part of circuit". The left, top, and bottom branches extend out to connect to unknown components. We don't know the potential difference between those outer endpoints!

We must find a contiguous path where all components and currents are known. The path from $a$ to $c$, down through the $1\Omega$ resistor to $d$, and left to $b$ is perfectly known. Let's apply Kirchhoff's Voltage Law (KVL) along this path:

1. From $a$ to $c$: We move with the $3\text{ A}$ current through the $5\Omega$ resistor. Potential drops by $3\times 5=15\text{ V}$. 2. From $c$ to $d$: We move with the $3\text{ A}$ current through the $1\Omega$ resistor. Potential drops by $3\times 1=3\text{ V}$. 3. From $d$ to $b$: We move against the $3\text{ A}$ current through the $2\Omega$ resistor. Potential rises by $3\times 2=6\text{ V}$.

Writing this as an equation:

The potential difference across the capacitor is $12\text{ V}$.

Finally, the energy $U$ stored in a capacitor is given by the elegant formula:

Substituting our known values ($C=4\mu \text{F}=4\times 10^{-6}\text{ F}$ and $V=12\text{ V}$):

To express this in millijoules (mJ), we shift the decimal point:

And there we have it! By carefully navigating the known parts of the circuit, we've successfully unlocked the energy stored within the capacitor.